Solution
= Solution
Apply the <Cauchy-Schwarz inequality> to $(x,y,z)$ and $(1,1,1)$ in the Euclidean <inner product> on $\mathbb R^3$. It gives
$$
(x+y+z)^2\leq3(x^2+y^2+z^2).
$$
After expansion and division by two,
$$
\boxed{x^2+y^2+z^2\geq xy+yz+zx.}
$$
The equality criterion from the introductory proof says $x=y=z$. One can also see both the inequality and equality case from $x^2+y^2+z^2-xy-yz-zx=\tfrac12[(x-y)^2+(y-z)^2+(z-x)^2]$.